EQUILIBRIUM OF FLEXIBLE CORDS 69
Taking the axes along the tangent and the normal through the middle point of the element and applying the conditions of equilibrium we obtain
3 (r+ dT) cos^ - Tcos ~ - F (- ds)* = 0,
2i 2
d& or dT
and Nds-2Tsm~-dTsin- = 0,
J-J <U
where dB is the angle between the two tensile forces which act at the ends of the element. But since the cord is supposed to be perfectly flexible the tensile forces are tangent to the surface of contact. Therefore 9 is the angle between the tangents, and consequently the angle between the normals, at the ends of the element. As an angle becomes indefinitely small its cosine approaches unity and its sine approaches the angle itself ,t therefore we can make the substitutions
do , , . dB de cos— = 1 and sin— = —
in the last two equations, and obtain
dT+Fds = 0, (1)
and Nds - TdO + § dT de = 0. (2)
Neglecting the differential of the second order in equation (2) and then eliminating ds between equations (1) and (2) we get
(3)
where M is the coefficient of friction. Integrating the last
* The negative sign in F ( — ds) indicates the fact that F and ds are measured in opposite directions. f See Appendix AVI.